Graph question (1 Viewer)

[BlueGas]

Can someone explain the answers for both parts of the question please?

 

[psyc1011]

(i) Increasing.

The sign of the derivative shows when the curve is increasing (>0) or decreasing (<0) because derivative is the gradient function. i.e. input a point and it outputs the gradient at the point to the curve or gradient of tangent.

Increasing: gradient always positive so curve is always rising.
Decreasing: gradient always negative so curve is always falling.

As you can see in the graph, f' is always positive, so gradient is always positive from x = 0 to x = infinity. Looks like it has a horizontal asymptote at y = 0 hence will never go below 0. So f is increasing.


(ii) Concave down.

The sign of the second derivative shows when curve is concave up or concave down. It is also the gradient function of f'.

Concave up: like a frown. Rise up then stop at stationary point, then fall down.
Concave down: like a smile. Fall down then stop at stationary point, then rise up.

From graph, f' always has negative gradient so we say f'' < 0 for all real x. Hence f'' is concave down.

 

[keepLooking]

i) By inspecting the f'(x) graph, you know that in the domain, f'(x) is always > 0, hence you know f(x) is increasing throughout its domain.

EDIT: psyc1011 has a better answer above ;D

 

[BlueGas]

(i) Increasing.

The sign of the derivative shows when the curve is increasing (>0) or decreasing (<0) because derivative is the gradient function. i.e. input a point and it outputs the gradient at the point to the curve or gradient of tangent.

As you can see in the graph, f' is always positive, so gradient is always positive from x = 0 to x = infinity. Looks like it has a horizontal asymptote at y = 0 hence will never go below 0. So f is increasing.

(ii) Concave down.

The sign of the second derivative shows when curve is concave up or concave down. It is also the gradient function of f'.

From graph, f' always has negative gradient so we say f'' < 0 for all real x. Hence f'' is concave down.

This is the part the confuses me. I've seen these types of graphs before and I know the the graph will never touch the x-axis, even though the curve is going down it'll never touch the x-axis so how is f(x) increasing?

 

[psyc1011]

This is the part the confuses me. I've seen these types of graphs before and I know the the graph will never touch the x-axis, even though the curve is going down it'll never touch the x-axis so how is f(x) increasing?

The graph displayed is f'(x), the gradient function. Assuming that it has a horizontal asymptote at y = 0, then it never touches the x-axis. That means f'(x) > 0, so always positive! Doesnt matter if the f' graph is swirling/jumping crazily as long as it is > 0 for all real x.

Positive gradient means it will always rise up and never fall down. i.e. increasing

 

[psyc1011]

The f'(x) is going down so it is increasing a decreasing rate

 

[psyc1011]

By inspecting the curve, the gradient function (f') itself is decreasing (starts off with high gradient then lowers down near 0), so we can say f is increasing at a decreasing rate.

Example values,

x = 0.01, y' = 100
x = 0.2, y' = 5
x = 5, y' = 0.2
x = 100, y' = 0.01

Since y' is clearly decreasing and always positive, then y' is increasing at a decreasing rate

 

[BlueGas]

The f'(x) is going down so it is increasing a decreasing rate

Yeah that's a much better way to word it, because the graph is actually increasing but the numbers get smaller, that's what confused me, so how about the second part, if it's basically a "bowl" shape it's always concave down and if it's a "hill" shape then it's concave up? Is it always like this?

 

[BlueGas]

How about if a question was referring to y = x^3, when f'(x) = 3x^2, so looking at the graph 3x^2 it seems that it's increasing so the curve must be increasing too? But when looking at x^3 it doesn't seem like it's increasing.

 

[psyc1011]

Yeah that's a much better way to word it, because the graph is actually increasing but the numbers get smaller, that's what confused me, so how about the second part, if it's basically a "bowl" shape it's always concave down and if it's a "hill" shape then it's concave up? Is it always like this?

Yes, always. But with respect to f(x), not f'(x).

 

[psyc1011]

How about if a question was referring to y = x^3, when f'(x) = 3x^2, so looking at the graph 3x^2 it seems that it's increasing so the curve must be increasing too? But when looking at x^3 it doesn't seem like it's increasing.

If f' is increasing, then f is increasing is a logical error because that is the incorrect relation. f' can be negative, hence f is not always increasing.

Example, y = x^2 and y' = 2x. y' is increasing on all x but y isn't increasing on all x.

As for f(x) = x^3,

f' is not increasing. It is decreasing from -inf to 0 then increases from 0 to +inf.

f' is positive for all real x hence f is increasing for all real x.

Look at y = x^3 and check it's gradient at all points. Just draw tangents at all points on the graph. Do you see that all the tangents have a positive gradient? Hence, increasing for all x.